For the function f(x) = -2x2 + 8, what is the name of its graph, its intercepts, axis of symmetry, and how to draw it?

The function $f(x) = -2x^2 + 8$ is a quadratic function of the form $ax^2 + bx + c$.
The graph of a quadratic function is a parabola.
The name of the graph is a parabola.

1.1.2. Calculate both the x-intercepts and y-intercepts of the graph above.

The x-intercepts are $\boxed{(-2, 0) \text{ and } (2, 0)}$.

The y-intercept is $\boxed{(0, 8)}$.

1.1.3. What is the axis of symmetry of the graph.

For $f(x) = -2x^2 + 8$, we have $a = -2$ and $b = 0$.

The axis of symmetry is $\boxed{x = 0}$.

1.1.4. Draw the graph of $f(x)$ and show the x-intercepts and the y-intercepts on it.
To draw the graph:

• Plot the x-intercepts at $(-2, 0)$ and $(2, 0)$.
• Plot the y-intercept at $(0, 8)$. This is also the vertex of the parabola since the axis of symmetry is $x=0$.
• Since the coefficient of $x^2$ is negative ($-2$), the parabola opens downwards.
• Draw a smooth curve connecting these points, opening downwards and symmetric about the y-axis ($x=0$).

1.2. The following diagram represents the graph of $f(x) = \frac{a}{x} + q$. A(4;2) is a point on the graph.
The question asks to determine the values of $a$ and $q$.

From the graph, the horizontal asymptote is the line that the graph approaches as $x$ tends to positive or negative infinity. This line is $y=1$.
For a function of the form $f(x) = \frac{a}{x} + q$, the horizontal asymptote is $y=q$.
Therefore, $q = 1$

Substitute $x=4$, $f(x)=2$, and $q=1$ into the equation $f(x) = \frac{a}{x} + q$.

The values are $\boxed{a = 4 \text{ and } q = 1}$.